$ABC$ is a plane lamina of the shape of an equilateral triangle. $D$, $E$ are mid-points of $AB$, $AC$ and $G$ is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through $G$ and perpendicular to the plane $ABC$ is $I_0$. If part $ADE$ is removed, the moment of inertia of the remaining part about the same axis is $\dfrac{NI_0}{16}$ where $N$ is an integer. Value of $N$ is ______.
Numerical value type. Enter your answer.
Answer: 11
For an equilateral lamina of mass $M$ and side $a$, $I_0 = \dfrac{Ma^{2}}{12}$.
Triangle $ADE$ has side $\dfrac a2$ and mass $\dfrac M4$, so about its own centroid $I' = \dfrac M4\cdot\dfrac{(a/2)^{2}}{12} = \dfrac{I_0}{16}$.
Its centroid is at height $\tfrac23h$ and $G$ at $\tfrac13h$ ($h = \tfrac{\sqrt3}{2}a$), so they are $\tfrac h3 = \dfrac{a}{2\sqrt3}$ apart. Parallel axes:
$$I_{ADE} = \frac{I_0}{16} + \frac M4\cdot\frac{a^{2}}{12} = \frac{I_0}{16} + \frac{I_0}{4} = \frac{5I_0}{16}$$
Remaining: $I_0 - \dfrac{5I_0}{16} = \dfrac{11I_0}{16}$, so $N = 11$.
Solution by Sreeraj P, M.Sc Physics